Упр.24.11 ГДЗ Мерзляк 10 класс Базовый уровень (Алгебра)

1) 1+sin(a)+cos(a)=2v2cos(a/2)cos(a/2-?/4);
2) (1+cos(4a-2?)+cos(4a-?/2))/(1+cos(4a+?)+cos(4a+3?/2))=ctg(2a);
3) sin^2(15?/8-2a)-cos^2(17?/8-2a)=-cos(4a)/v2.

1. $$1+\sin a+\cos a=1+2\sin\frac a2\cos\frac a2+\left(2\cos^2\frac a2-1\right)$$

   $$=2\cos\frac a2\left(\sin\frac a2+\cos\frac a2\right)=2\cos\frac a2\left(\sin\frac a2+\sin\left(\frac\pi2-\frac a2\right)\right)$$

   $$=2\cos\frac a2\cdot 2\sin\frac{\frac\pi2-\frac a2+\frac a2}{2}\cos\frac{\frac\pi2-\frac a2-\frac a2}{2}$$

   $$=4\cos\frac a2\sin\frac\pi4\cos\left(\frac\pi4-\frac a2\right)=2\sqrt2\cos\frac a2\cos\left(\frac a2-\frac\pi4\right).$$

2. $$\frac{1+\cos(4a-2\pi)+\cos\left(4a-\frac\pi2\right)}{1+\cos(4a+\pi)+\cos\left(4a+\frac{3\pi}{2}\right)}$$

   $$=\frac{1+\cos4a+\sin4a}{1-\cos4a+\sin4a}$$

   $$=\frac{1+\left(2\cos^2 2a-1\right)+\sin4a}{1-\left(1-2\sin^2 2a\right)+\sin4a} =\frac{2\cos^2 2a+2\sin2a\cos2a}{2\sin^2 2a+2\sin2a\cos2a}$$

   $$=\frac{2\cos2a(\cos2a+\sin2a)}{2\sin2a(\sin2a+\cos2a)} =\frac{\cos2a}{\sin2a}\cdot\frac{\cos2a+\cos\left(\frac\pi2-2a\right)}{\sin2a+\sin\left(\frac\pi2-2a\right)}$$

   $$=\frac{\cos2a}{\sin2a}\cdot \frac{2\cos\frac{\frac\pi2-2a+2a}{2}\cos\frac{\frac\pi2-2a-2a}{2}} {2\sin\frac{\frac\pi2-2a+2a}{2}\cos\frac{\frac\pi2-2a-2a}{2}}$$

   $$=\frac{\cos2a}{\sin2a}\cdot\frac{\cos\frac\pi4}{\sin\frac\pi4}=\ctg2a.$$

3. $$\sin^2\left(\frac{15\pi}{8}-2a\right)-\cos^2\left(\frac{17\pi}{8}-2a\right)$$

   $$=\sin^2\left(\frac{15\pi}{8}-2a\right)-\cos^2\left(\frac\pi2+\frac{13\pi}{8}-2a\right)$$

   $$=\sin^2\left(\frac{15\pi}{8}-2a\right)-\sin^2\left(\frac{13\pi}{8}-2a\right)$$

   $$=\left(\sin\left(\frac{15\pi}{8}-2a\right)-\sin\left(\frac{13\pi}{8}-2a\right)\right) \left(\sin\left(\frac{15\pi}{8}-2a\right)+\sin\left(\frac{13\pi}{8}-2a\right)\right)$$

   $$=2\sin\frac\pi8\cos\left(\frac{7\pi}{8}-2a\right)\cdot 2\sin\frac{7\pi}{8}\cos\frac\pi8$$

   $$=\frac1{\sqrt2}\sin\left(\frac{3\pi}{2}-4a\right)=-\frac{\cos4a}{\sqrt2}.$$

Ответ

$$1)\;1+\sin a+\cos a=2\sqrt2\cos\frac a2\cos\left(\frac a2-\frac\pi4\right);$$
$$2)\;\frac{1+\cos(4a-2\pi)+\cos\left(4a-\frac\pi2\right)}{1+\cos(4a+\pi)+\cos\left(4a+\frac{3\pi}{2}\right)}=\ctg2a;$$
$$3)\;\sin^2\left(\frac{15\pi}{8}-2a\right)-\cos^2\left(\frac{17\pi}{8}-2a\right)=-\frac{\cos4a}{\sqrt2}.$$